hdu 1080(LCS变形)

Human Gene Functions

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3008    Accepted Submission(s): 1701

Problem Description

It
is well known that a human gene can be considered as a sequence,
consisting of four nucleotides, which are simply denoted by four
letters, A, C, G, and T. Biologists have been interested in identifying
human genes and determining their functions, because these can be used
to diagnose human diseases and to design new drugs for them.

A
human gene can be identified through a series of time-consuming
biological experiments, often with the help of computer programs. Once a
sequence of a gene is obtained, the next job is to determine its
function. One of the methods for biologists to use in determining the
function of a new gene sequence that they have just identified is to
search a database with the new gene as a query. The database to be
searched stores many gene sequences and their functions – many
researchers have been submitting their genes and functions to the
database and the database is freely accessible through the Internet.

A
database search will return a list of gene sequences from the database
that are similar to the query gene. Biologists assume that sequence
similarity often implies functional similarity. So, the function of the
new gene might be one of the functions that the genes from the list
have. To exactly determine which one is the right one another series of
biological experiments will be needed.

Your job is to make a
program that compares two genes and determines their similarity as
explained below. Your program may be used as a part of the database
search if you can provide an efficient one.

Given two genes
AGTGATG and GTTAG, how similar are they? One of the methods to measure
the similarity of two genes is called alignment. In an alignment, spaces
are inserted, if necessary, in appropriate positions of the genes to
make them equally long and score the resulting genes according to a
scoring matrix.

For example, one space is inserted into AGTGATG
to result in AGTGAT-G, and three spaces are inserted into GTTAG to
result in –GT--TAG. A space is denoted by a minus sign (-). The two
genes are now of equal length. These two strings are aligned:

AGTGAT-G
-GT--TAG

In
this alignment, there are four matches, namely, G in the second
position, T in the third, T in the sixth, and G in the eighth. Each pair
of aligned characters is assigned a score according to the following
scoring matrix.

* denotes that a space-space match is not allowed. The score of the alignment above is (-3)+5+5+(-2)+(-3)+5+(-3)+5=9.

Of
course, many other alignments are possible. One is shown below (a
different number of spaces are inserted into different positions):

AGTGATG
-GTTA-G

This
alignment gives a score of (-3)+5+5+(-2)+5+(-1) +5=14. So, this one is
better than the previous one. As a matter of fact, this one is optimal
since no other alignment can have a higher score. So, it is said that
the similarity of the two genes is 14.

 

Input

The
input consists of T test cases. The number of test cases ) (T is given
in the first line of the input. Each test case consists of two lines:
each line contains an integer, the length of a gene, followed by a gene
sequence. The length of each gene sequence is at least one and does not
exceed 100.

 

Output

The output should print the similarity of each test case, one per line.

 

Sample Input

2
7 AGTGATG
5 GTTAG
7 AGCTATT
9 AGCTTTAAA

 

Sample Output

14
21

 

Source

 

题意：给你两个串，问在哪些位置添加多少个-号可以使权值最大。

 

分析：和LCS差不多吧。。分三个状态考虑。。下午训练时，状态方程出来了初始化一直没想到一直没出答案，无心再想这个题了。。今天先记录下吧。。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#define N 105
using namespace std;

int mp[][]=
{
    {,-,-,-,-},
    {-,,-,-,-},
    {-,-,,-,-},
    {-,-,-,,-},
    {-,-,-,-,}
};
int dp[N][N];
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        int n,m;
        char str1[],str2[];
        scanf("%d%s",&n,str1+);
        scanf("%d%s",&m,str2+);
        memset(dp,,sizeof(dp));
        int x,y;
        for(int i=; i<=n; i++) ///这里很重要
        {
            if(str1[i]=='A') y=;
            if(str1[i]=='C') y=;
            if(str1[i]=='G') y=;
            if(str1[i]=='T') y=;
            dp[i][] = dp[i-][] + mp[y][];
        }
        for(int i=; i<=m; i++)
        {
            if(str2[i]=='A') x=;
            if(str2[i]=='C') x=;
            if(str2[i]=='G') x=;
            if(str2[i]=='T') x=;
            dp[][i] = dp[][i-] + mp[][x];
        }

        for(int i=; i<=n; i++)
        {
            for(int j=; j<=m; j++)
            {

                    if(str1[i]=='A') x=;
                    if(str1[i]=='C') x=;
                    if(str1[i]=='G') x=;
                    if(str1[i]=='T') x=;
                    if(str2[j]=='A') y=;
                    if(str2[j]=='C') y=;
                    if(str2[j]=='G') y=;
                    if(str2[j]=='T') y=;
                    dp[i][j] = max(dp[i-][j-]+mp[x][y],max(dp[i-][j]+mp[][x],dp[i][j-]+mp[][y]));
            }
        }
        //for(int i=1;i<=n;i++)
        printf("%d\n",dp[n][m]);
    }
    return ;
}